MAXIMA MINIMA AND POINTS OF INFLECTION
MAXIMA, MINIMA AND POINTS OF INFLECTION.
Relative Maximum points
Relative Minimum points
Maxima and minima are also called TURNING points or STATIONARY points.
Consider the gradient of this curve at points along the curve:
Grad = 0 Grad = – ¼
Grad = – ½ Grad = 1
Grad = ¼
Grad = – ¼ Grad = ½
Grad = ½ P
Grad = ¼
Grad = 0
Grad = 1
Going from left to right, the gradient is DECREASING up to the point P and then it starts INCREASING again.
The point P where the gradient stops decreasing and starts increasing is called an INFLECTION point.
These are ordinary inflection points:
Grad decreasing
Grad decreasing
infl infl
point point
Grad increasing Grad increasing
This is a special inflection point because the gradient is also zero.
It is called a STATIONARY INFLECTION POINT.
Grad increasing
Stationary inflection point
Grad decreasing
(A typical example of this is the curve y = x3)
The safest test for a Maximum or a Minimum or a Stationary Inflection point is the 1st Derivative Test.
We simply find the gradient on either side of the point where the gradient is zero.
eg For the curve y = x(x – 3)2 = x(x2 – 6x + 9) = x3 – 6x2 + 9x
dy = 3x2 – 12x + 9
dx
= 3(x2 – 4x + 3)
= 3(x – 1)(x – 3) = 0 at max/min (or point of inflection
So gradient is zero at x = 1 and x = 3
1st derivative test :
|x values |0 |1 |2 | |2 |3 |4 |
|Gradient |+9 |0 |– 3 | |– 3 |0 |+9 |
MIN
MAX at x = 3
at x = 1
This is a sketch of the graph: (1, 4)
(3, 0)
1 2 3
To understand the 2nd Derivative Test, consider these curves:
y
y = x(x – 3)2 MAX
INFL
= x3 – 6x2 + 9x
(cubic curve) MIN
1 2 3 4
y ′
y ′ = 3x2 – 12x + 9
= 3(x2 – 4x + 3)
= 3(x – 1)(x – 3) 1 2 3 4
(parabola)
y ′′
y ′′ = 6x – 12
(line graph)
1 2 3 4
NOTICE THREE POINTS:
When the cubic has a MAXIMUM the 2nd derivative is a NEGATIVE number.
When the cubic has a MINIMUM the 2nd derivative is a POSITIVE number.
When the cubic has an INFLECTION point the 2nd derivative is ZERO.
Unfortunately, the test does not always work!
Sometimes if d2y = 0 it can be a MAX or a MIN instead of a point of inflection.
dx2
y = x4
eg if y = x4
dy = 4x3 which equals zero if x = 0
dx
and d2y = 12x2 = 0 when x = 0 (implying a point of inflection at x = 0)
dx2
Clearly y = x4 has a MIN at x = 0
However if we use the 1st derivative test it is clearly a minimum point:
|x values |– 1 |0 |1 |
|gradients |– 4 |0 |4 |
Min point
In general, for a curve y = f(x) the MAX or MIN points are where dy = 0
dx
and we can decide whether it is a MAX or a MIN using the 2nd derivative test which
basically says : if d2y < 0 it is a MAX and if d2y > 0 it is a MIN
dx2 dx2
Usually there is a point of inflection when d2y = 0
dx2
1. Consider the curve y = x2(6 – x ) = 6x2 – x3
dy = 12x – 3x2
dx
= 3x(4 – x) = 0 at max/min
So x = 0 or 4
d2y = 12 – 6x
dx2
2nd derivative test : At x = 0 d2y = 12 so MIN At x = 4 d2y = –12 so MAX
dx2 dx2
Point of inflection is when d2y = 0
dx2
(4, 32)
so 12 – 6x = 0
x = 2
1 2 3 4 5 6
Alternatively, using the 1st derivative test: dy = 12x – 3x2 = 3x(4 – x)
dx
| x values |–1 |0 |1 | |3 |4 |5 |
|gradient |–15 |0 |9 | |9 |0 |–15 |
Min(0, 0) Max(4, 32)
In the following example, the 2nd derivative test fails:
2. Consider the curve y = x4 – 4x3 = x3(x – 4)
y′ = 4x3 – 12x2
= 4x2(x – 3) = 0 for stationary points (max/min/infl)
So x = 0 and 3
y′′ = 12x2 – 24x = 12x(x – 2)
2nd derivative test:
We must now resort to the 1st derivative test at x = 0:
This is a sketch of the curve :
Stationary Infl (0, 0)
1 2 3 4 5
Infl (2, –16 ) Min (3, – 27)
3. Find any Max/Min/Infl points on the curve y = xex
y′ = xex + 1ex = 0 at max/min/infl
so ex(x + 1) = 0
ex ≠ 0 so x = – 1
Using 1st derivative test :
|x values |– 2 |– 1 |0 |
|gradient |e– 2(– 1) |0 |e0(1) |
| |= – | |= +1 |
Min
y′′ = xex +ex + ex = 0 at infl pt
Min (– 1, – 0.37)
ex(x + 2) = 0
x = –2, y = – 0.27
____________________________________________________________________
4. Find any Max/Min/Infl points on the curve y = x2 = x + 1 + 1 .
(x – 1) (x – 1)
This graph has a vertical asymptote at x = 1 and an oblique one y = x + 1
y′ = (x – 1)2x – x2 = 0 at turning pts
(x – 1)2
2x2 – 2x – x2 = 0
x2 – 2x = 0
x(x – 2) = 0
x = 0 and 2
1st deriv. Test
|x values |– 1 |0 |½ | |1 ½ |2 |3 |
|grad |+ |0 |– | |– |0 |+ |
Max Min
(0, 0) (2, 4)
| | | |
| y = ax2 + bx + c | | |
| |1 | |
|or = (x – p)(x – q) | | |
|y = ax3 + bx2 + cx + d | | |
| |2 | |
|or = (x – p)(x – q)(x – r) | | |
|y = ax4 + bx3 + …… | | |
| |3 | |
A stationary inflection point counts as 2 turning points combined.
eg. y = x3
Sometimes the above “rule” fails if the gradient cannot equal zero.
eg. y = x3 – 3x2 + 9x = x(x2 – 3x + 9) which crosses the x axis only at x = 0
y′ = 3x2 – 6x + 9 = 0 at max/min
so 3(x2 – 2x + 3) = 0
which has no real solutions 7
so the gradient cannot equal zero.
The point of inflection can be found:
y′′ = 6x – 6 = 0 at infl pt. 0 1 2
so x = 1 y = 7
NB the gradient decreases as we move from x = 0 to x = 1 then it starts to increase again from x = 1 onwards. y′ = 3(x – 1)2 + 6
|x |0 |.5 |.9 |1 |1.1 |1.5 |2 |
|y′ |9 |6.75 |6.03 |6 |6.03 |6.75 |9 |
CONCAVITY.
Concave Concave
down up
Concave down when d2y < 0 Concave up when d2y > 0
dx2 dx2
(hint: think “decelerating”) (hint: think “accelerating”)
Clearly a curve changes its curvature at points of inflection.(see diagram below)
1. Find the max/min points and the range of values of x for which the following
curve is concave down.
y = (x + 3)2(x – 3)2 = (x + 3)(x – 3)(x + 3)(x – 3) = (x2 – 9)2 = x4 – 18x2 + 81
y′ = 4x3 – 36x = 0 at max/min
4x(x2 – 9) = 0
x = 0, ±3
y′′ = 12x2 – 36 = 0 at inflection points i.e. x = ±√3
2nd deriv. test:
|x = – 3 |x = 0 |x = +3 |
|y′′ = + |y′′ = – |y′′ = + |
| MIN |MAX |MIN |
ꞌ
concave up concave up
concave down
–3 –√3 √3 3
Concave down if d2y > 0 so 12x2 – 36 > 0 ie x2 > 3
dx2
The curve is CONCAVE DOWN for –√ 3 < x < √ 3
2. Find the range of values of x for which the following curve is concave upwards.
y = x2(x – 12) = x3 – 12x2
y′ = 3x2 – 24x = 0 at max/min
3x(x – 8) = 0
x = 0 and 8
2nd deriv. test :
|x = 0 |x = 8 |
|y′′ = – |y′′ = + |
|MAX |MIN |
4 8 12
Concave upwards if d2y > 0
dx2
6x – 24 > 0
6x > 24
This curve is concave upwards for x > 4 ( or 4 < x < ∞ )
3. A function is defined parametrically as follows:
x = t3 + 24t y = t2 – 2t
(a) Find the x, y coordinates of the turning point of this function.
(b) Find the values of t, x and y when this function is concave upwards.
(a) dx = 3t2 + 24 dy = 2t – 2
dt dt
dy = dy × dt = 2t – 2 = 0 at max/min point
dx dt dx 3t2 + 24
2t = 2
t = 1 so x = 25 and y = –1
(b) d2y = dt × d dy = dt × d 2t – 2
dt2 dx dt dx dx dt 3t2 + 24
= 1 × (3t2 + 24)2 – (2t – 2)(6t)
3t2 + 24 (3t2 + 24)2
= 6t2 + 48 – 12t2 + 12t
(3t2 + 24)3
= –6t2 + 12t + 48
(3t2 + 24)3
= –6( t2 – 2t – 8)
(3t2 + 24)3
= –6(t – 4)(t + 2) = 0 at inflection points so t = 4, –2
(3t2 + 24)3
By 2nd derive test, the value of y ′′ at t = 1 is Positive so (25, –1) is a min
The function must be concave upwards between points of inflection
ie for –2 < t < 4 between ( –56, 8) and (160, 8)
|t |– 2 |4 |
|x |– 56 |160 |
|y |8 |8 |
Min point (25, –1)
– 56 160
4. For the curve given parametrically as x = 4t3 and y = 3t2 – 7t find the range of
values of t for which the curve is concave upwards.
dx = 12t2 and dy = 6t – 7
dt dt
so dy = dy × dt
dx dt dx
dy = 6t – 7 ( NB grad is zero when t = 7 x =6.35, y = – 4)
dx 12t2 6
d2y = d dy = dt × d 6t – 7
dx2 dx dx dx dt 12t2
= 1 × 12t2 × 6 – (6t – 7)24t
12t2 (12t2)2
= 72t2 – 144t2 + 168t
(12t2)3
= 168t – 72t2
(12t2)3
= 12t ( 14 – 6t)
(12t2)3 (NB curve is vertical at t = 0 , grad is ∞ )
Inflection points are where d2y = 0 ie at t = 0 and 7
dx2 3
This is concave up when d2y > 0 so 0 < t < 7
dx2 3
(NB the x values for which the curve is concave up are 0 < x < 50.8)
You would not be expected to produce this type of graph.
Curve is vertical at this inflection point. Grad = ∞
2nd inflection point
50.8
Min at (6.35, – 4)
-----------------------
y ′′ > 0
y ′′ = 0
y ′′< 0
At x = 3 y′′ = 12x2 – 24x
= 36 so MIN at x = 3
At x = 0 y′′ = 12x2 – 24x
= 0 ( neither + nor – )
So we cannot say whether max or min.
Also, inflection points are where y′′ = 0
12x(x – 2) = 0
So x = 0 and 2
We must now resort to the 1st derivative test at x = 0:
|x values |–1 |0 |1 |
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Stationary
Infl point
You could just say that the curve is concave down if –√ 3 < x < √ 3 because there is a Maximum point in between or if yꞌꞌ > 0.
y′′ = 6x – 24 = 0 at inflection point
x = 4
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have
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